$ g(x) = \int_{\,2}^{\,x} {\frac{13}{{1 + {t^{\,2}}}}\,dt}$ $ g\,^\prime(5)=$
Explanation: The Fundamental Theorem of Calculus If $~ g(x)=\int_a^xf(t)\,dt\,$, then $~g^\prime (x)=f(x)\,$ This only works if $f$ is continuous on $[a,b]$. Thankfully, the function $f(t) = \dfrac{13}{{1 + {t^{\,2}}}}$ is continuous on $[2,5]$. Applying the theorem We're given: $ g(x) = \int_{2}^{x} {\frac{13}{{1 + {t^{\,2}}}}\,dt} $ So the theorem tells us: $ g\,^\prime(x) = \frac{13}{{1 + {x^{\,2}}}} $ Evaluating $g'(5)$ $ g\,^\prime(5) =\frac{13}{{1 + {x^{\,2}}}}= \frac{13}{{1 + 5^2}}=\dfrac{13}{26}=\dfrac12 $ The answer: $g'(5)=\dfrac{1}{2}$